Question

我试图编写一个接受字符串的方法，然后查找至少包含两个元音的单词，并将它们添加到另一个数组中，如下所示：

anyString = "Today we had football for breakfast"
#=> "today", "football", "breakfast"

如果这是另一种语言，我会做几个for循环，然后在计数器达到两个以上时添加单词。我想知道在Ruby中最简单的方法是什么。我试图在算法方面做得更好。有人可以提出什么建议吗？

Answer 1

"Today we had football for breakfast"
.downcase.scan(/\w*[aeiou]\w*[aeiou]\w*/)
#=> ["today", "football", "breakfast"]

如果你想将其概括为＆＃34;至少n个元音，＆＃34;然后用

替换正则表达式

/(?:\w*[aeiou]){n}\w*/

Answer 2

我会使用String#count方法。

str = "Today we had football for breakfast"
str.downcase.split.select { |s| s.count("aeiou") > 1 }
# => ["today", "football", "breakfast"]

Answer 3

你正在考虑像C这样的命令式语言。这将是一种更为Rubyesque的方式：

anyString.downcase.split.select {|word| word.scan(/[aeiou]/).size >= 2 }

Answer 4

str = "Today we had football for breakfast"

str.downcase.split.select { |w| w.gsub(/[^aeiou]/,'').size > 1 }
  #=> ["today", "football", "breakfast"]

我使用了downcase而不是(/[^aeiou]/i，因为问题（"Today"）中给出的示例表明要收集缩小的字词。

Answer 5

其他变种

anyString.downcase.split(' ').select { |word| word.match(/.*[aeiou].*[aeiuo].*/) }

Answer 6

any_string.split.select { |w| w.scan(/[aeiou]/i).size > 1 }
=> ["Today", "football", "breakfast"]

简单的Ruby方法

6 个答案: